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you have a horizontal grindstone (a disk) that is 87 kg, has a 0.35 m radius, is turning at 95 rpm (in the positive direction), and you press a steel axe against the edge with a force of 15 n in the radial direction. (a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many turns will the stone make before coming to rest?

Sagot :

varshamittal029

The angular acceleration is 0.19 and the number of turns is 41.07 revolutions.

m = 87 kg be the mass of the grindstone,

r = 0.35 m be the radius of the grindstone,

I = mr²/2 be the grindstone's moment of inertia,

ω = 95 revolution/min = 95 2π/60s = 9.9 s-1 be initial angular velocity,

F = 15 N be the force pressing radially,

f = 0.2 be the coefficient of friction,

α (unknown) be the grindstones angular acceleration.

a)The tangential force slowing the grindstone is

fF

Its torque is

rfF

By Newton's Second Law

rfF = Iα

Therefore

α = rfF/I = 2rfF/mr² = 2fF/mr

Substituting actual numbers

α = 2×0.2×15/(87×0.35) = 0.19 s-2

Note that the above is the absolute value of the acceleration.

Depending on which direction is considered positive or negative,

the answer may need to be negated.

b)

Let

t be the time it takes the grindstone to come to rest,

θ be the total angle by which the grindstone will turn before coming to rest.We have the following identities, which follow from the definition of acceleration:

ω = αt

θ = αt²/2

From the first

t = ω/α

Plug into the second

θ = α(ω/α)²/2 = ω²/2α

Substitute actual numbers

θ = 9.9²/2×0.19 = 257.92

The above result is in radians.

Since the result is supposed to be given in revolutions,

we need to divide by the number of radians per revolutions:

257.92/2π = 41.07 revolutions

Therefore, the angular acceleration is 0.19 and the number of turns is 41.07 revolutions.

To learn more about torque refer here

https://brainly.com/question/20691242

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